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The problem presented involves a normal distribution of rice bag weights. Here's a breakdown:

- The weights are normally distributed with mean \(\mu\) and standard deviation \(\sigma\).
- **97.5%** of the bags weigh more than **30.5 kg**.
- **88.49%** of the bags weigh less than **46.3 kg**.
- We are tasked with finding the values of \(\mu\) and \(\sigma\).

### Step 1: Use of Z-scores
From the standard normal distribution table (or a calculator), we can convert the given percentages into Z-scores:

1. **97.5% of the bags weigh more than 30.5 kg**:
   - This means **2.5%** weigh less than **30.5 kg**.
   - From the Z-table, the Z-score corresponding to **2.5%** in the lower tail is approximately **-1.96**.

2. **88.49% of the bags weigh less than 46.3 kg**:
   - From the Z-table, the Z-score corresponding to **88.49%** is approximately **1.19**.

### Step 2: Set up the equations
Using the Z-score formula:  
\[
Z = \frac{X - \mu}{\sigma}
\]
where \(X\) is the data value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

- For the **30.5 kg** bag:  
\[
-1.96 = \frac{30.5 - \mu}{\sigma}
\]
This gives the equation:
\[
30.5 - \mu = -1.96\sigma \quad \text{(Equation 1)}
\]
or
\[
\mu = 30.5 + 1.96\sigma \quad \text{(Equation 1')}
\]

- For the **46.3 kg** bag:  
\[
1.19 = \frac{46.3 - \mu}{\sigma}
\]
This gives the equation:
\[
46.3 - \mu = 1.19\sigma \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations
We now solve the two equations:

1. \(\mu = 30.5 + 1.96\sigma\)
2. \(46.3 - \mu = 1.19\sigma\)

Substitute Equation 1 into Equation 2:
\[
46.3 - (30.5 + 1.96\sigma) = 1.19\sigma
\]
Simplify:
\[
46.3 - 30.5 - 1.96\sigma = 1.19\sigma
\]
\[
15.8 = 3.15\sigma
\]
\[
\sigma = \frac{15.8}{3.15} \approx 5.02
\]

### Step 4: Find \(\mu\)
Substitute \(\sigma = 5.02\) into Equation 1:
\[
\mu = 30.5 + 1.96 \times 5.02 \approx 30.5 + 9.84 = 40.34
\]

### Final Answer:
- \(\mu \approx 40.34 \, \text{kg}\)
- \(\sigma \approx 5.02 \, \text{kg}\)

Would you like further details or clarifications?

---

### Follow-up Questions:
1. What is the significance of the Z-score in the context of this problem?
2. How do we interpret the mean and standard deviation in real-life scenarios?
3. Can you explain why the Z-score table was used and its importance?
4. What would happen if we changed the percentage from 97.5% to 99%?
5. How would you solve this problem if only one percentage was given?

#### Tip:
Always ensure you know how to read and use the Z-table for solving normal distribution problems like this one.

The mass of a bag of rice, in kg, is normally distributed with mean μ and standard deviation σ. If 97.5% of the bags of rice have masses of more than 30.5 kg and 88.49% of the bags of rice have masses of less than 46.3 kg, find the value of μ and σ.

This problem involves calculating the mean (μ) and standard deviation (σ) for a normally distributed dataset. Using the Z-score formula, we derive the values based on given percentages of rice bags weighing more than 30.5 kg and less than 46.3 kg. This is ideal for students learning about normal distribution in statistics.

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